Topics Covered Discrete probability distributions –The Uniform Distribution –The Binomial Distribution –The Poisson Distribution Each is appropriately

Topics Covered

• Discrete probability distributions

– The Uniform Distribution

– The Binomial Distribution

– The Poisson Distribution

• Each is appropriately applied in certain situations and to particular phenomena

Continuous Random Variables

• Continuous random variable can assume all real

number values within an interval (e.g., rainfall, pH)

• Some random variables that are technically discrete

exhibit such a tremendous range of values, that is

it desirable to treat them as if they were continuous

variables, e.g. population

• Continuous random variables are described by

probability density functions

• Probability density functions are defined using the same rules required of probability mass functions, with some additional requirements:

• The function must have a non-negative value throughout the interval a to b, i.e.

f(x) >= 0 for a <= x <= b• The area under the curve defined by f(x), within

the interval a to b, must equal 1

Probability Density Functions

x

area=1f(x)

a b

The Normal Distribution

• The most common probability distribution is the normal distribution

• The normal distribution is a continuous distribution that is symmetric and bell-shaped

Source: http://mathworld.wolfram.com/NormalDistribution.html


• Most naturally occurring variables are distributed normally (e.g. heights, weights, annual temperature variations, test scores, IQ scores, etc.)

• A normal distribution can also be produced by tracking the errors made in repeated measurements of the same thing; Karl Friedrich Gauss was a 19th century astronomer who found that the distribution of the repeated errors of determining the position of the same star formed a normal (or Gaussian) distribution

• The probability density function of the normal

distribution:

• You can see how the value of the distribution at x

is a f(x) of the mean and standard deviation


2)(5.0

2

1)(

x

exf

Source: http://en.wikipedia.org/wiki/Normal_distribution

λ

The Poisson Distribution & The Normal Distribution

The probability density function of a normal distribution approximating the probability mass function of a binomial distribution

Source: http://en.wikipedia.org/wiki/Normal_Distribution


• As with all frequency distributions, the area under the curve between any two x values corresponds to the probability of obtaining an x value in that range

• The total area under the curve is equal to one

Source: http://en.wikipedia.org/wiki/Normal_Distribution


• The way we use the normal distribution is to find the probability of a continuous random variable falling within a range of values ([c, d])

The probability of the variable between c and d

is the area under the curve

• The areas under normal curves are given in tables such as that found in Table A.2 in Appendix A. Textbook (Rogerson)

x

f(x)

c d

Normal Tables

• Variables with normal distributions may have an infinite number of possible means and standard deviations

Source: http://en.wikipedia.org/wiki/Normal_distribution

Normal Tables

• Variables with normal distributions may have an infinite number of possible means and standard deviations

• Normal tables are standardized (a standard normal distribution with a mean of zero and a standard deviation of one)

• Before using a normal table, we must transform our data to a standardized normal distribution

Standardization of Normal Distributions

• The standardization is achieved by converting the data into z-scores

s

xxi z-score

• The z-score is the means that is used to transform our normal distribution into a standard normal distribution ( = 0 & = 1)

StandardizationMonth T (°F) Z-score

J 39.53 -1.56

F 46.36 -1.03

M 46.42 -1.02

A 60.32 0.05

M 66.34 0.51

J 75.49 1.22

J 75.39 1.21

A 77.29 1.36

S 68.64 0.69

O 57.57 -0.16

N 54.88 -0.37

D 48.2 -0.89

Mean = 59.70Standard deviation = 12.97

Original data: Z-score: Mean = 0Standard deviation = 1

• Example I – A data set with = 55 and = 20:

• Using z-scores in conjunction with standard normal tables (like Table A.2 on page 214 of Rogerson) we can look up areas under the curve associated with intervals, and thus probabilities (P(X>1.75) or P(X<1.75))

x - Z-score = = x - 5520

• If one of our data values x = 90 then:

x - Z-score = = 90 - 5520

= 3520

= 1.75


Look Up Standard Normal Tables

• Using our example z-score of 1.75, we find the position of 1.75 in the table and use the value found there


μ = 0

f(x)

+1.75

.0401 P(Z > 1.75) = 0.0401

P(Z <= 1.75) = 0.9599

μ = 55

f(x).0401

+90

P(X > 90) = 0.0401

P(X <= 90) = 0.9599

• Example II – If we have a data set with = 55 and = 20, we calculate z-scores using:

• Using z-scores in conjunction with standard normal tables (like Table A.2 on page 214 of Rogerson) we can look up areas under the curve associated with intervals, and thus probabilities (P(X>20) or P(X<=20))

x - Z-score = = x - 5520

• If one of our data values x = 20 then:

x - Z-score = = 20 - 5520

= -3520

= -1.75


μ = 0

f(x)

+1.75-1.75

Look Up Standard Normal Tables• Using our example z-score of -1.75, we find the position of 1.75 in the table and use the value found there; because the normal distribution is symmetric the table does not need to repeat positive and negative values


μ = 0

f(x)

+1.75-1.75

4.01% (.0401)4.01% (.0401)


μ = 0

f(x)

μ = 55

f(x)

c d

-1.75

4.01% (.0401)

4.01% (.0401)

P(Z <= -1.75) = 0.0401

P(Z > -1.75) = 0.9599

P(X <= 20) = 0.0401

P(X > 20) = 0.9599

Finding the P(x) for Various Intervals

a

3.

a

2.

a

1.

P(0 Z a) = [0.5 – (table value)]•Total Area under the curve = 1, thus the area above x is equal to 0.5, and we subtract the area of the tail

P(Z a) = [1 – (table value)]•Total Area under the curve = 1, and we subtract the area of the tail

P(Z a) = (table value)• Table gives the value of P(x) in the tail above a

Finding the P(x) for Various Intervals

a

6.

a

5.

a

4.

P(Z a) = [1 – (table value)]• This is equivalent to P(Z a) when a is positive

P(Z a) = (table value)• Table gives the value of P(x) in the tail below a, equivalent to P(Z a) when a is positive

P(a Z 0) = [0.5 – (table value)]• This is equivalent to P(0 Z a) when a is positive

Finding the P(x) for Various IntervalsP(a Z b) if a < 0 and b > 0

a

7. b

• With this set of building blocks, you should be able to calculate the probability for any interval using a standard normal table

or= [0.5 – (table value for a)] + [0.5 – (table value for b)]= [1 – {(table value for a) + (table value for b)}]

= (0.5 – P(Z<a)) + (0.5 – P(Z>b))= 1 – P(Z<a) – P(Z>b)

Finding the P(x) – Example

• Suppose we are in charge of buying stock for a shoe store. We will assume that the distribution of shoe sizes is normally distributed for a gender. Let’s say the mean of women’s shoe sizes is 20 cm and the standard deviation is 5 cm

• If our store sells 300 pairs of a popular style each week, we can make some projections about how many pairs we need to stock of a given size, assuming shoes fit feet that are +/- 0.5 cm of the length of the shoe (because of course we must use intervals here!)

Finding the P(x) – Example• 1. How many pairs should we stock of the 25

cm size?• We first need to convert this to a range

according to the assumption specified above, since we can only evaluate P(x) over an interval P(24.5 cm ≤ x ≤ 25.5 cm) is what we need to find

• Now we need to convert the bounds of our interval into z-scores:

x - Zlower = = 24.5 - 205

= 4.55

= 0.9

x - Zupper = = 25.5 - 205

= 5.55

= 1.1


• 1. How many pairs should we stock of the 25 cm size?

• We now have our interval expressed in terms of z-scores as P(0.9 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables

1.1

0.9

• We can calculate this area by finding P(0.9Z 1.1) = P(Z0.9) - P(Z1.1)

Finding the P(x) – Example• 1. How many pairs should we stock of the 25

cm size?• We now have our interval expressed in terms of

z-scores as P(0.9 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables

1.1

0.9 • We can calculate this area by finding P(0.9Z 1.1) = P(Z0.9) - P(Z1.1) = 0.1841 – 0.1357 = 0.0484

• Now multiply our P(0.9 ≤ Z ≤ 1.1) by total sales per week to get the number of shoes we should stock in that size: 300 х 0.0484 = 14.52 ≈ 15

Finding the P(x) – Example• 2. How many pairs should we stock of the 18 cm

size? (assuming shoes fit feet that are +/- 0.5 cm of the length of the

shoe) • Again, we convert this to a range according to the

assumption specified above, since we can only evaluate P(x) over an interval P(17.5 cm ≤ x ≤ 18.5 cm) is what we need to find

• Now we need to convert the bounds of our interval into z-scores:

x - Zlower = = 17.5 - 205

= -2.55

= -0.5

x - Zupper = = 18.5 - 205

= -1.55

= -0.3


• 2. How many pairs should we stock of the 18 cm size?

• We now have our interval expressed in terms of z-scores as P(-0.5 ≤ Z ≤ -0.3) for use in a standard normal distribution, which we can evaluate using our standard normal tables

-0.5-0.3

• We can calculate this area by finding P(-0.5Z-0.3) = P(Z-0.3) - P(Z-0.5)

Finding the P(x) – Example• 2. How many pairs should we stock of the 18 cm

size?• We now have our interval expressed in terms of z-

scores as P(-0.5 ≤ Z ≤ -0.3) for use in a standard normal distribution, which we can evaluate using our standard normal tables

-0.5-0.3 • We can calculate this area by finding

P(-0.5Z-0.3) = P(Z-0.3) - P(Z-0.5) = 0.3821 – 0.3085 = 0.0736• Now multiply our P(-0.5 ≤ Z ≤ -0.3) by total sales

per week to get the number of shoes we should stock in that size: 300 х 0.0736 = 22.08 ≈ 22


• 3. How many pairs should we stock in the 18 to 25 cm size range?

• As always, we convert this to a range according to the assumption specified above, since we can only evaluate P(x) over an interval P(17.5 cm ≤ x ≤ 25.5 cm) is what we need to find

• We have already found the appropriate Z-scores

x - Zlower = = 17.5 - 205

= -2.55

= -0.5

x - Zupper = = 25.5 - 205

= 5.55

= 1.1

Finding the P(x) – Example• 3. How many pairs should we stock in the 18 to

25 cm size range?• We now have our interval expressed in terms of z-

scores as P(-0.5 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables

-0.51.1

• We can calculate this area by finding P(-0.5Z1.1) = 1 – [P(Z-0.5) + P(Z1.1)] = 1 – [0.3085 + 0.1357] = 1 – 0.4442 = 0.5558• Now multiply our P(-0.5 ≤ Z ≤ 1.1) by total sales

per week to get the # of shoes we should stock in that range: 300 х 0.5558 = 166.74 ≈ 167

Commonly Used Probabilities

-3σ -2σ -1σ μ +1σ +2σ +3σ

f(x)

68%

95%

99.7%

-σ μ +σ

f(x)


1)(

xz

• Standard normal distribution:

P(z > 1) = 0.1587 p(-1 ≤ z ≤ 1) = 1 – 2*0.1587 = 0.6826

• Normal distribution as illustrated in the Figure:

P(μ-σ ≤ x ≤ μ-σ) = 0.6826 ≈ 0.68

• For a normal distribution, 68% of the observations lie within about one standard deviations of the mean

-2σ μ +2σ

f(x)


2)2(

xz


P(z > 2) = 0.0228 p(-2 ≤ z ≤ 2) = 1 – 2*0.0228 = 0.9543


P(μ-2σ ≤ x ≤ μ-2σ) = 0.9543 ≈ 0.95

• For a normal distribution, 95% of the observations lie within about two standard deviations of the mean

-3σ μ +2σ

f(x)


3)3(

xz


P(z > 3) = 0.0013 p(-3 ≤ z ≤ 3) = 1 – 2*0.0013 = 0.9974


P(μ-3σ ≤ x ≤ μ-3σ) = 0.9974 ≈ 0.997

• For a normal distribution, 99.7% of the observations lie within about two standard deviations of the mean


-3σ -2σ -1σ μ +1σ +2σ +3σ

f(x)

68%

95%

99.7%

Documents

Topics Covered Discrete probability distributions –The Uniform Distribution –The Binomial Distribution –The Poisson Distribution Each is appropriately